Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

There are 96 total possibilities (i.e, the denominator).

There are 12 numbers 96/8 that are multiples of/divisible by 8 themselves.

If n = 6, then you have 6*(6+1)(6+2) or 6*7*8, so that number will be divisible by 8. Same for n = 14. 14(14+1)(14+2), 14*15*16...also divisible by 8 because 16 is divisible by 8. For every number that is divisible by 8, (every 8th number to 96) will have 3 values for n that create a number with n(n+1)(n+2) that will be divisible by 8, so I multiplied 12 by 3 = 36/96

36/96 = 6 / 16 = 3 / 8

EDIT: I just did this in Excel and came up with 5/8 (60/96). I went wrong because I didn't consider that n=2 will have 2*3*4, or 8*3, which will be divisible by 8 also!

What's the best way to figure this out?

My answer to my own question:

We know that any set of number n(n+1)(n+2) that will be divisible by 2, three separate times ( like 16 /2 = 8, 8/2 = 4, 4/2 = 2 is divisible by 2 three times) will be divisible by 8. Since we know that 2 + 3 + 4 works 2 /2 , and 4/ 2, = 2 ...2/2 = 3 times... every even number will satisfy this question, so 48 of them.

Then we also know that any odd number that includes a multiple of 8 as (n+1) [ n+2 doesn't matter be odd + 2 = odd and will never be divisble by 8] will also satisfy this. There is only 1 odd number that satisfies this for every multiple of 8 up to 96. We've already established there 96/8 = 12, so there are 12 odd numbers that also satisfy this question. so 48 + 12 = 60. Probability = 60/96 or 5/8

tarek99 wrote:

If an integer n is to be chosen at random from the integers 1 to 96, inclusive, what is the probability that n(n+1)(n+2) will be divisible by 8?

a) 1/4 b) 3/8 c) 1/2 d) 5/8 e) 3/4

Any fast way to solve this???

_________________

------------------------------------ J Allen Morris **I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.

If an integer n is to be chosen at random from the integers 1 to 96, inclusive, what is the probability that n(n+1)(n+2) will be divisible by 8?

a) 1/4 b) 3/8 c) 1/2 d) 5/8 e) 3/4

Any fast way to solve this???

D, what is difficulty of this problem? took me about 3-4 minutes to solve. I broke down #s but I think there is a simpler way.

(1-10), 4,5,6,7,8,10 were the only ones that worked.

6*10= 60/96

Found it wasnt as difficult as it was finding out which digits worked...

First, 5 does not work. 5*6*7 = 210 210 / 8 = 26.25.

Second, 2 does work. 2*3*4 = 24, 24/8 = 3.

Also, where did you get " *10 " ? if you have 6 possible for 1-10, and you are saying you have ten "groups" of 10, that's incorrect. The numbers go from 1 - 96. There can't possibly be 10 groups of 10 in 96 numbers. _________________

------------------------------------ J Allen Morris **I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.

If an integer n is to be chosen at random from the integers 1 to 96, inclusive, what is the probability that n(n+1)(n+2) will be divisible by 8?

a) 1/4 b) 3/8 c) 1/2 d) 5/8 e) 3/4

Any fast way to solve this???

D, what is difficulty of this problem? took me about 3-4 minutes to solve. I broke down #s but I think there is a simpler way.

(1-10), 4,5,6,7,8,10 were the only ones that worked.

6*10= 60/96

Found it wasnt as difficult as it was finding out which digits worked...

First, 5 does not work. 5*6*7 = 210 210 / 8 = 26.25.

Second, 2 does work. 2*3*4 = 24, 24/8 = 3.

Also, where did you get " *10 " ? if you have 6 possible for 1-10, and you are saying you have ten "groups" of 10, that's incorrect. The numbers go from 1 - 96. There can't possibly be 10 groups of 10 in 96 numbers.

At second glance, I have no idea how I came up with that #. Pure luck.

Attacking this problem again and having major issues.... There is a pattern but taking to long to uncover.

At second glance, I have no idea how I came up with that #. Pure luck.

Attacking this problem again and having major issues.... There is a pattern but taking to long to uncover.

help?

gmatnub had it right -- just didn't explain it fully.

all even numbers should qualify because the product of consecutive even numbers is always divisible by 8. This is because one of the consecutive even numbers will be divisible by 2 and the other will be divisible by 4.

that gets us to 48 of the 96 numbers.

now, you have to count the odd numbers such that n+1 is divisible by 8. This is the case with all x-1 such that x is divisible by 8.

96/8 = 12 numbers b/w 1 and 96.

(48+12)/96 = 5/8

gmatclubot

Re: PS: Probability
[#permalink]
24 Jul 2008, 13:47